Thermodynamics of the Universe.
This document suggests a model of the universe not well accepted by mainstream scientists. Do not copy it as you will get an 'F' on your term paper as well as be ridiculed by friends.
Let us consider the Friedmann equation for a flat universe
(eq 1) (D′/D)2 = (8πG/3)ε / c2
Where G is Newton's gravitational constant, c is the speed of light, ε is energy density, D is distance between stars and D’ is the rate expansion over time. This equation adequately describes some instances where objects in the cosmos seem to recess from each others. Our goal here is to reinterpret this and other equations frequently used in studies of the universe to provide an alternate view, which is mathematically consistent and sensible with observable data.
Our model assumes the expansion is an illusion from measuring the universe with a ruler that has been slowly shrinking over centuries. The measuring stick consists of ruler units. A dimension characterizes each unit. We tentatively describe the unit as a sphere and the dimension as its radius (denoted as letter a). We define
D = Aa
Where A is some constant. The expansion rate of the universe relates to the contraction rate of the ruler unit as
D’ = -Aa’
(eq 2) D’/D = -a’/a
(eq 3) (a′/a)2 = (8πG/3)ε / c2
This reinterpretation, as is, is simply a change of perspective. We will next turn our attention to the first law of thermodynamics, on which the expansion model relies to hypothesize a dark energy fueling the expansion. Symbolically, the law can be stated as:
E’ + PV’ = TS’
Where E is the internal energy of the gas, P is pressure, V is volume, T is temperature, and S is a quantity called entropy, which is a measure of the microstates present in the system. TS’ is also equal to the flow of heat into or out of the given volume.
In the expansion universe model, the first law is applied to the universe and the process is said to be adiabatic. As there is no net heat flow, TS’ = 0. Therefore we have the simpler case E’ + PdV = 0, or PV′ = -E’. This implies a negative force energizing the expansion of the universe.
In the shrinking ruler model, we use the first law of thermodynamics to model a ruler unit. This unit contracts hence its volume change V’ is negative. We will investigate the special case when the unit does not change internal energy as it shrinks, therefore
(eq 4) E’ = 0
PV’ = heat flow out
We indirectly compute heat loss of the ruler unit by estimating the entropy change of the universe as the unit shrinks.
Entropy (S) of a system is related to the number of microstates (W) as:
S = k ln W
Where k is Boltzmann’s constant and W is the number of microstates associated with a particular macroscopic state.
Assume W is inversely proportional to the volume of the ruler unit. The smaller the unit, the more microstates there are in the universe. This is probably not exact but will be used to explore concepts.
(eq 5) W = U/V
Where U is a constant and has the unit of volume.
S = k ln W = k ln(U/V)
S = k ln U - k ln V
The time derivative of the above is
S’ = (k ln U)’ – (k ln V)’
Since derivative of a constant is 0, we have
S‘ = – (k ln V)’ = -kV’/V
S’ has the unit of energy/(time*temperature). As heat flow out to the universe is equal in magnitude to the heat loss by the ruler unit, substituting the negative of the above approximation into the first law of thermodynamics for a shrinking ruler unit gives
PV’ = TkV’/V
(eq 6) PV = kT
Let us call this the first shrinking ruler equation.
Consider the ruler unit as a sphere (radius = a) in comoving coordinates, the mass-energy inside the sphere is E = εV. The density ε is (like a) a function of time, so the derivative E′ = (4π/3)(a3ε′ + 3a2a′ε). Since we assume internal energy of the ruler unit does not change (eq 4), we have
E′ = (4π/3)(a3ε′ + 3a2a′ε) = 0
Eliminating 4π/3 and moving 3a2a′ε to the right hand side give
a3ε′ = -3a2a′ε
Dividing both sides by εa3 gives
(eq 7) ε′ / ε = -3a′ /a
Let us call this the second shrinking ruler equation.
We now return to Friedmann’s equation for a flat universe, applied to the shrinking ruler (eq 3).
(a′/a)2 = (8πG/3)ε / c2
Taking another derivative of the Friedmann equation with ε gives
2(a′/a)[(aa″-a′2)/a2] = (8πG/3) ε′ / c2
Using the second ruler equation (eq 7), we can eliminate ε′ and get
2(a′/a)[(aa″-a′2)/a2] = -(8πG/3) 3(a′/a) ε / c2
Dividing both sides by 2(a′/a) and noting that 3/3 =1 on the right hand side gives
(aa″-a′2)/a2 = -(4πG) ε / c2
Expanding the above yields
a″/a -a′2/a2 = -(4πG) ε / c2
Adding (a′/a)2 gives
a″/a = -(4πG) ε / c2 + (a′/a)2
Substituting (a′/a)2 from the Friedmann equation gives
a″/a = -(4πG) ε / c2 + (8πG/3) ε / c2
Finally, by simplifying the right hand side, we have
(eq 8) a″/a = -(4πG / 3 ) ε / c2
This is the acceleration equation of the shrinking ruler. This tells us the old gravitation force is energizing the clumping of the universe. By inverting the view and working with an expanding universe model, scientists have to define a negative energy to doubly negate gravity.
Taking the Hubble constant H = -a’/a = 71.5/30.9×1018 = 2.3×10-18 s−1, we have
(a′/a)2 = (8πG/3)ε / c2 = (-2.3×10-18)2 = 5.29-36 s-2
Comparing the above with the acceleration equation, we can conclude:
a″/a = -(4πG / 3 ) ε / c2 = -2.65×10-36 s-2
Astro-physicists have invested considerable resources and reputation in the expanding universe model, too much to change. But, time does wonder and a future Einstein may turn things upside down. Until then, do see this article as off-the-beaten-track but not off-course. It is licensed under a Creative Commons Attribution 3.0 Unported License. You are free to distribute, remix, tweak and build upon this work as long as credit is given to CreationWord.com.
I am not a career astrophysicist. This and other articles on the universe stem from my attempts to reach the Creator via examining the creation. I hope you find something useful as you read them judiciously. Thanks. D. N. Pham.
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